By Freeman J Dyson; David Derbes

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Also ∇ · jN = − ∂ρH ∆E =i ρN ∂t (119) The electrostatic potential V of the nucleus has the matrix element given by ∇2 V = −4πρN (120) The states being spherically symmetric, ρ N is a function of r only, and so the general solution of Poisson’s equation simplifies to 12 V (r) = − 6π r r 0 r12 ρN (r1 ) dr1 (121) Outside the nucleus V (r) = Ze2 /r is constant in time, and so the matrix element of V (r) for this transition is zero. In fact from (119) and (120) we get by integration V (r) = i∆E (−4π)(−r)jN o (r) = 4πr jN o (r) i∆E (122) where jN o is the outward component of the current.

Thus a continuum state given by the spinor ψ = u exp {(ip · x − iEt)/ } is normalized so that |E| (104) u∗ u = mc2 Now if we multiply the Dirac equation for a free particle, (44), by u on the left, we get Eu∗ βu = cu∗ βα · pu + mc2 u∗ u; its complex conjugate is Eu∗ βu = −cu∗ βα · pu + mc2 u∗ u since βα is anti-Hermitian; then by adding we get (105) E uu = mc2 u∗ u Therefore the normalization becomes uu = +1 for electron states = −1 for positron states = ; This is the definition of . (106) With this normalization the density of states in momentum space is one per volume h3 of phase space, that is to say ρ= 1 mc2 dp1 dp2 dp3 h3 |E| (107) per volume dp1 dp2 dp3 of momentum space, for each direction of spin and each sign of charge.

E = density of final states per unit energy interval. VBA is the matrix element of the potential V for the transition. Here V may be anything, and may be itself a second-order or higher order effect obtained by using higher-order perturbation theory. The difficulties in real calculations usually come from the factors 2 and π and the correct normalization of states. Always I shall normalize the continuum states not in the usual way (one particle per unit volume) which is non-invariant, but instead One particle per volume 31 mc2 |E| (103) 32 Advanced Quantum Mechanics where |E| is the energy of the particles.